URGENT! Need help with probability question (1 Viewer)

Danneo · Oct 25, 2018

Thank god if someone replies as HSC math is within a few hours

-If 3% of coins produced are faulty, what is the probability that a sample of 3 coins taken from the production line will have atleast two faulty coins

I keep getting the quesiton wrong

Ans = 1323/500000 (I think thats the right ans)

InteGrand · Oct 25, 2018

Danneo said:
Thank god if someone replies as HSC math is within a few hours

-If 3% of coins produced are faulty, what is the probability that a sample of 3 coins taken from the production line will have atleast two faulty coins

I keep getting the quesiton wrong

Ans = 1323/500000 (I think thats the right ans)

$\noindent This is a question of \textbf{binomial probability} (though I'm not sure the 2U syllabus calls it that). The probability of any single coin being faulty is $p = 0.03$ (i.e. $3\%$).$

$\noindent To have at least 2 coins in the sample of 3 coins be faulty, this means either exactly 2 or exactly 3 coins are faulty. So the answer is equal to $\color{blue}{P(\text{exactly 2 coins faulty})}\color{black} + \color{red}{P(\text{exactly 3 coins faulty})}$.$

$\noindent Note that $\color{blue}{P(\text{exactly 2 coins faulty}) = 3p^2 (1-p)}$. Also, $\color{red}{P(\text{exactly 3 coins faulty}) = p^{3}}$.$

$\noindent Therefore, the answer is $ \color{blue}{3p^2 (1-p)} \color{black}+ \color{red}{p^{3}}\color{black} = 3\times (0.03)^{2} \times (1-0.03) + (0.03)^{3} = \frac{1323}{500000} (= 0.002646)$.$

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$\noindent The way to calculate $\color{blue}{P(\text{exactly 2 coins faulty})}$ is to note that there are $3$ orders our sample could result in having two coins faulty ($F$) and hence one coin not faulty ($N$); these orders are $FFN,FNF,NFF$ (i.e. the non-faulty coin could be the third coin we picked, or the second, or the first). The probability of each of these cases is $p\times p\times (1-p) = p^{2}(1-p)$ (since $p$ is the probability for a given coin to be faulty and $1-p$ is the probability of a given coin being not faulty), and since there are $\color{magenta}{3}$ orders (each equally likely), the probability is $\color{blue}{P(\text{exactly 2 coins faulty})} \color{black}= \color{magenta}{3}\color{black}p^{2}(1-p)$.$

$\noindent In general, the probability a sample of $n$ coins has exactly $k$ faulty ones would be $\binom{n}{k}p^{k}(1-p)^{n-k}$. If you haven't learnt about binomial coefficients $\binom{n}{k}$ (sometimes written $^{n}C_{k}$) in the course, then don't worry about this.$

HeroWise · Oct 25, 2018

Imagine this comes in the HSC

Danneo · Oct 26, 2018

Holy crap, this just changed how i see binomial, so just tested 1 - (P(No faults) + (1 fault))
P(No faults) = (97/100)^3
P(Exactly 1 fault) = 3C1*(3/100)*(97/100)^2
and it worked, thank you good sir

URGENT! Need help with probability question (1 Viewer)

Danneo

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InteGrand

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HeroWise

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Danneo

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